I am self-studying differential equations using MIT's publicly available materials. If $P$ is a monic polynomial of degree $n$ and if $y$ is a solution to the differential equation $p(D)y = 0$ (where $D$ is the differential operator) and we have initial conditions
\begin{align} y(0) = 1 \qquad y'(0) = y''(0) = \cdots = y^{(n-1)}(0) = 0, \end{align} we are asked to show that if
a) $P(0) = 0$, we can discover nothing from $\mathcal{L}[y](s) = Y(s)$ about the coefficients of $p$; but if
b) $P(0) \neq 0$, then $P$ is uniquely determined by $Y(s)$.
I feel like I'm teetering on the brink of a solution, but somehow I can't quite fall into it. Here's what I've done:
Let $p = x^n + a_1x^{n-1} + \cdots + a_n$. It is evident that \begin{align} \mathcal{L}[y^{(j)}](s) & = s^jY(s) - s^{(j-1)} \end{align} for $j = 1, \ldots, n$. So taking the Laplace transform of both sides of our differential equation, we have \begin{align} (s^nY(s) - s^{(n-1)}) + a_1(s^{(n-1)}Y(s) -s^{(n-2)}) + \cdots + a_{n-1}(sY(s) - 1) + a_nY(s) & = 0\Rightarrow\\ Y(s)(s^n + a_1s^{n-1} + \cdots + a_n) - (s^{n-1} + a_1s^{n-2} + \cdots + a_{n-1}) & = 0\\ P(s)Y(s) - (s^{n-1} + a_1s^{n-2} + \cdots + a_{n-1}) & = 0. \end{align} And setting $s = 0$, we get \begin{align} P(0)Y(0) & = a_{n-1}\Rightarrow\\ a_{n-1} & = 0. \end{align} if $P(0) = 0$. This seems wrong, for consider the case $n = 1$. $P(0) = 0$ means we in fact have $P(x) = x$ -- that is, $a_{n-1} = a_0 = 1$.
No doubt I'm missing something elementary here--I'd appreciate some help figuring out exactly what.
With credit to Semiclassical and user1952009, we can continue from where I stopped to obtain
\begin{align} P(s)Y(s) - \frac{P(s) + a_n}{s} & = 0\Rightarrow\\ P(s)(sY(s) - 1) - a_n & = 0\Rightarrow\\ P(s) & = \frac{P(0)}{sY(s) - 1} \end{align}
which shows that $P(s)$ is completely determined by $Y(s)$ unless $P(0) = 0$, because if $P(0) = 0$, \begin{align} Y(s) = \frac{P(s) + P(0)}{sP(s)} \end{align} which is undefined at $s = 0$.